Right here:)))))))))))))))))
Answer:
A
Step-by-step explanation:
Recall the double angle identity:
![\sin2\theta=2\sin\theta\cos\theta](https://tex.z-dn.net/?f=%5Csin2%5Ctheta%3D2%5Csin%5Ctheta%5Ccos%5Ctheta)
With
measuring between 0º and 90º, we know
. So from the Pythagorean identity, we get
![\sin^2\theta+\cos^2\theta=1\implies\cos\theta=\sqrt{1-\sin^2\theta}=\dfrac{\sqrt{21}}5](https://tex.z-dn.net/?f=%5Csin%5E2%5Ctheta%2B%5Ccos%5E2%5Ctheta%3D1%5Cimplies%5Ccos%5Ctheta%3D%5Csqrt%7B1-%5Csin%5E2%5Ctheta%7D%3D%5Cdfrac%7B%5Csqrt%7B21%7D%7D5)
Then
![\sin2\theta=2\dfrac25\dfrac{\sqrt{21}}5=\dfrac{4\sqrt{21}}{25}](https://tex.z-dn.net/?f=%5Csin2%5Ctheta%3D2%5Cdfrac25%5Cdfrac%7B%5Csqrt%7B21%7D%7D5%3D%5Cdfrac%7B4%5Csqrt%7B21%7D%7D%7B25%7D)
Answer:
![P(X](https://tex.z-dn.net/?f=P%28X%20%3C1.96%29%20%3D%200.975)
![P(X >1.64) = 0.0505](https://tex.z-dn.net/?f=P%28X%20%3E1.64%29%20%3D%200.0505)
![P(0.5 < X < 0.5) = 0](https://tex.z-dn.net/?f=P%280.5%20%3C%20X%20%3C%200.5%29%20%3D%200)
Step-by-step explanation:
Given
--- Mean
--- Variance
Calculate the standard deviation
![\sigma^2 = 1](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D%201)
![\sigma = 1](https://tex.z-dn.net/?f=%5Csigma%20%3D%201)
Solving (a): P(X < 1.96)
First, we calculate the z score using:
![z = \frac{X - \bar x}{\sigma}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cbar%20x%7D%7B%5Csigma%7D)
This gives:
![z = \frac{1.96 - 0}{1}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B1.96%20-%200%7D%7B1%7D)
![z = \frac{1.96}{1}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B1.96%7D%7B1%7D)
![z = 1.96](https://tex.z-dn.net/?f=z%20%3D%201.96)
The probability is then solved using:
![(X < 1.96) = P(z](https://tex.z-dn.net/?f=%28X%20%3C%201.96%29%20%3D%20P%28z%20%3C1.96%29)
From the standard normal distribution table
![P(z](https://tex.z-dn.net/?f=P%28z%20%3C1.96%29%20%3D%200.97500)
So:
![P(X](https://tex.z-dn.net/?f=P%28X%20%3C1.96%29%20%3D%200.975)
Solving (b): P(X > 1.64)
First, we calculate the z score using:
![z = \frac{X - \bar x}{\sigma}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cbar%20x%7D%7B%5Csigma%7D)
This gives:
![z = \frac{1.64 - 0}{1}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B1.64%20-%200%7D%7B1%7D)
![z = \frac{1.64}{1}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B1.64%7D%7B1%7D)
![z = 1.64](https://tex.z-dn.net/?f=z%20%3D%201.64)
The probability is then solved using:
![(X > 1.64) = P(z >1.64)](https://tex.z-dn.net/?f=%28X%20%3E%201.64%29%20%3D%20P%28z%20%3E1.64%29)
![P(z >1.64) = 1 - P(z](https://tex.z-dn.net/?f=P%28z%20%3E1.64%29%20%3D%201%20-%20P%28z%3C1.64%29)
From the standard normal distribution table
![P(z >1.64) = 1 - 0.9495](https://tex.z-dn.net/?f=P%28z%20%3E1.64%29%20%3D%201%20-%200.9495)
So:
![P(X >1.64) = 0.0505](https://tex.z-dn.net/?f=P%28X%20%3E1.64%29%20%3D%200.0505)
Solving (c): P(0.5 < X < 0.5)
This can be split as:
![P(0.5 < X < 0.5) = P(0.5](https://tex.z-dn.net/?f=P%280.5%20%3C%20X%20%3C%200.5%29%20%3D%20P%280.5%3CX%29%20-%20P%28X%3C0.5%29)
In probability:
![P(0.50.5)](https://tex.z-dn.net/?f=P%280.5%3CX%29%20%3D%201%20-%20P%28X%3E0.5%29)
![P(0.5](https://tex.z-dn.net/?f=P%280.5%3CX%29%20%3D%201%20-%20%5B1%20-%20P%28X%3C0.5%29%5D)
![P(0.5](https://tex.z-dn.net/?f=P%280.5%3CX%29%20%3D%201%20-%201%20%2B%20P%28X%3C0.5%29)
![P(0.5](https://tex.z-dn.net/?f=P%280.5%3CX%29%20%3D%20%20P%28X%3C0.5%29)
becomes
![P(0.5 < X < 0.5) = P(X](https://tex.z-dn.net/?f=P%280.5%20%3C%20X%20%3C%200.5%29%20%3D%20P%28X%3C0.5%29%20-%20P%28X%3C0.5%29)
![P(0.5 < X < 0.5) = 0](https://tex.z-dn.net/?f=P%280.5%20%3C%20X%20%3C%200.5%29%20%3D%200)
Answer:
The answer to number 1 is -2 and the answer to number 2 is 3.
Step-by-step explanation:
Hope that helps!